Example Problem - Toroid - Revision history

Kirkbetz at 23:41, 20 January 2010

2010-01-20T23:41:49Z

← Older revision		Revision as of 16:41, 20 January 2010
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	== Points for page ==		== Points for page ==
	Worth		Worth 117

			Kirk total for paper 117 + 50 conference = 167

Kirkbetz at 23:39, 20 January 2010

2010-01-20T23:39:37Z

← Older revision		Revision as of 16:39, 20 January 2010
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	== Read by ==		== Read by ==

			== Points for page ==
			Worth

Kirkbetz at 06:10, 20 January 2010

2010-01-20T06:10:49Z

← Older revision		Revision as of 23:10, 19 January 2010
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	Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.		Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.

	==		== Reviewed by ==
	Reviewed by ==

	[[Will Griffith]]		[[Will Griffith]]
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	[[Matthew Fetke]]		[[Matthew Fetke]]

	==		== Read by ==
	Read by ==

Kirkbetz at 06:09, 20 January 2010

2010-01-20T06:09:00Z

← Older revision		Revision as of 23:09, 19 January 2010
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	[[Will Griffith]]		[[Will Griffith]]


	[[Matthew Fetke]]		[[Matthew Fetke]]

Kirkbetz at 06:08, 20 January 2010

2010-01-20T06:08:53Z

← Older revision		Revision as of 23:08, 19 January 2010
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	[[Will Griffith]]		[[Will Griffith]]


	[[Matthew Fetke]]		[[Matthew Fetke]]

	==		==
	Read by ==		Read by ==

Kirkbetz at 06:08, 20 January 2010

2010-01-20T06:08:23Z

← Older revision		Revision as of 23:08, 19 January 2010
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	Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.		Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.

	Reviewed ~~By:~~ Will Griffith		==
			Reviewed by ==

			[[Will Griffith]]
			[[Matthew Fetke]]

			==
			Read by ==

Dr. Ganon at 20:04, 19 January 2010

2010-01-19T20:04:35Z

← Older revision		Revision as of 13:04, 19 January 2010
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			By: Kirk Betz

	'''Problem:'''		'''Problem:'''
	Concerning Ampere's law		Concerning Ampere's law
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	Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.		Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.

			Reviewed By: Will Griffith

Kirkbetz at 19:58, 19 January 2010

2010-01-19T19:58:42Z

← Older revision		Revision as of 12:58, 19 January 2010
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	<math>r_m=(\frac{1}{2})\frac{OD + ID}{2} = 2.25\ cm</math>		<math>r_m=(\frac{1}{2})\frac{OD + ID}{2} = 2.25\ cm</math>

	Using the mean radius the mean path of length l_m can be calculated.		Using the mean radius the mean path of length <math>l_m</math> can be calculated.


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	<math>H_m=\frac{20 x 2.5}{.141}= 354.6\ A /m)</math>		<math>H_m=\frac{20 x 2.5}{.141}= 354.6\ A /m)</math>

	Since the width of the toroid is much smaller than the mean radius r_m we can assume a uniform H_m throughout teh cross-section of the toroid.		Since the width of the toroid is much smaller than the mean radius <math>r_m</math> we can assume a uniform <math>H_m</math> throughout teh cross-section of the toroid.

Kirkbetz at 19:54, 19 January 2010

2010-01-19T19:54:26Z

← Older revision		Revision as of 12:54, 19 January 2010
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	[[Image:toroid.jpg]]		[[Image:toroid.jpg]]


			''Figure created by Kirk Betz''



	'''Solution:'''		'''Solution:'''

Kirkbetz at 05:39, 19 January 2010

2010-01-19T05:39:40Z

← Older revision		Revision as of 22:39, 18 January 2010
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	'''Problem:'''		'''Problem:'''
			Concerning Ampere's law
	Let look at a coil around a toroid shown in the figure below. The coil has N = 20 turns around the toroid. The toroid has an inside diameter of ID = 4 cm and an outside diameter OD = 5 cm. Determine the field intensity H along the mean path length within the toroid with a current i = 2.5 A.		Let look at a coil around a toroid shown in the figure below. The coil has N = 20 turns around the toroid. The toroid has an inside diameter of ID = 4 cm and an outside diameter OD = 5 cm. Determine the field intensity H along the mean path length within the toroid with a current i = 2.5 A.

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	'''Solution:'''		'''Solution:'''

	Do symmetry the magnetic field intensity Hm along a circular contour within the toroid is constant. We can find the radius by		Do symmetry the magnetic field intensity Hm along a circular contour within the toroid is constant. We can find the mean radius by


	<math>r_m=(\frac{1}{2})\frac{OD + ID}{2}</math>		<math>r_m=(\frac{1}{2})\frac{OD + ID}{2} = 2.25\ cm</math>

			Using the mean radius the mean path of length l_m can be calculated.


	<math> l_m = { 2 \pi r_m} = 0.141\ </math>		<math> l_m = { 2 \pi r_m} = 0.141\ </math>

			With Ampere's Law (below) the field intensity along the mean path can be Found.

	<math>H_m=(\frac{Ni}{l_m})</math>		<math>H_m=(\frac{Ni}{l_m})</math>

			Finally teh H_m can be calculated.

	<math>H_m=\frac{20 x 2.5}{.141}= 354.6\ A /m)</math>		<math>H_m=\frac{20 x 2.5}{.141}= 354.6\ A /m)</math>

			Since the width of the toroid is much smaller than the mean radius r_m we can assume a uniform H_m throughout teh cross-section of the toroid.